All categories

3 years ago

A lightbulb is labeled "100 watts." what does "100 watts" represent?

2 answers:

6 0

That "100 watts" represent the power of the light-bulb. As it's power is 100 watts, it will glow shiner than 50 watts bulb, and dim than 200 watts of bulb

Charra [1.4K]3 years ago

3 0

That "100 watts" represent the power of the light-bulb. As it's power is 100 watts, it will glow shiner than 50 watts bulb, and dim than 200 watts of bulb

Hope this helps!

You might be interested in

Evelyn learns that a sound wave can be recorded electronically as an analog signal or as a digital signal. She investigates thes

Olin [163]

Answer:

I got the answer of yr question

but it is too long

brainly is not taking

5 0

3 years ago

Read 2 more answers

A long straight wire carries current to the right of the page. A rectangular loop is positioned directly under the wire in the p

slavikrds [6]

Answer:

Correct option A.

The net force exerted by this loop on the straight wire with the current is directed TOWARDS THE LOOP

Explanation:

The magnetic field exerts a force on a current-carrying wire in a direction given by the right-hand rule 1 (the same direction as that on the individual moving charges). This force can easily be large enough to move the wire since typical currents consist of very large numbers of moving charges.

Given that,

The wire's current is directed towards the right of the page.

The rectangular loops carry current in a clockwise direction.

Since the 'dot' field is increasing hence the induced magnetic field is 'cross', i.e. into the page and by the right-hand rule, the induced current is clockwise.

Then the magnetic field is into the page.

Since was known that

F= iL×B

Note the current is through the wire. Then, the length is in direction of the current.

Note: this equation gives the magnetic force that acts on a length L of a straight wire carrying a current (i) and immersed in a uniform magnetic field (B), that is perpendicular to the wire.

So, the magnetic field is always perpendicular to the current.

So using right hand rule,

F = i(L×B)

The length is to the right i.e. +x direction and the Magnetic field is perpendicular to the plane, i.e. in the +z direction

F = i (L•i × B•k)

F = iLB (i×k)

F = iLB•(-j)

F = -iLB•j

Then, the force is in the negative y-direction i.e. towards the loop.

8 0

4 years ago

The plates of a parallel plate capacitor are separated by d = 1.6 cm. The potential of the negative plate is 0 V, and the potent

yaroslaw [1]

Answer:

The electric field between the plates is 1875 V/m.

Explanation:

Given that,

Separated d=1.6 cm

Potential of negative plate = 0 V

Potential halfway between the plates = 15 V

We need to calculate the total potential

Using formula of potential

$V=2\times\text{Potential halfway between the plates}$

Put the value into the formula

$V=2\times15$

$V=30\ V$

We need to calculate the electric field between the plates

Using formula of electric field

$E=\dfrac{V}{d}$

$E=\dfrac{30}{1.6\times10^{-2}}$

$E=1875\ V/m$

Direction is negative as the field always points from positive to negative plate

Hence, The electric field between the plates is 1875 V/m.

5 0

3 years ago

What is the difference between a mineral, glass, and a rock?

IrinaVladis [17]

Answer:

A mineral is a solid formation that occurs in earth. A rock is a combination of more than just one mineral. And glass have more of a crystalline structure

Explanation:

3 0

2 years ago

Is it possible to have a charge of 5 x 10-20 C? Why?

ruslelena [56]

1) No

2) Yes

3) No

4) Equal and opposite

5) 32400 N

6) Repulsive

7) The electric force is $2.3\cdot 10^{39}$ times bigger than the gravitational force

Explanation:

In nature, the minimum possible charge that an object can have is the charge of the electron, which is called fundamental charge:

$e=1.6\cdot 10^{-19}C$

Electrons are indivisible particles (they cannot be separated), this means that an object can have at least the charge equal to the charge of one electron (in fact, it cannot have a charge less than $e$ , because it would meant that the object has a "fractional number" of electrons).

In this problem, the object has a charge of

$Q=5\cdot 10^{-20}C$

If we compare this value to $e$ , we notice that $Q$ , so no object can have a charge of $Q$ .

As we said in part 1), an object should have an integer number of electrons in order to be charged.

This means that the charge of an object must be an integer multiple of the fundamental charge, so we can write it as:

$Q=ne$

where

Q is the charge of the object

n is an integer multiple

e is the fundamental charge

Here we have

$Q=2.4\cdot 10^{-18}C$

Substituting the value of e, we find n:

$n=\frac{Q}{e}=\frac{2.4\cdot 10^{-18}}{1.6\cdot 10^{-19}}=15$

n is integer, so this value of the charge is possible.

We now do the same procedure for the new object in this part, which has a charge of

$Q=2.0\cdot 10^{-19}C$

Again, the charge on this object can be written as

$Q=ne$

where

n is the number of electrons in the object

Using the value of the fundamental charge,

$e=1.6\cdot 10^{-19}C$

We find:

$n=\frac{Q}{e}=\frac{2.0\cdot 10^{-19}}{1.6\cdot 10^{-19}}=1.25$

n is not integer, so this value of charge is not possible, since an object cannot have a fractional number of electrons.

To solve this part, we use Newton's third law of motion, which states that:

"When an object A exerts a force on an object B (Action force), then object B exerts an equal and opposite force on object A (reaction force)".

In this problem, we have two objects:

- A charge Q

- A charge 5Q

Charge Q exerts an electric force on charge 5Q, and we can call this action force. At the same time, charge 5Q exerts an electric force on charge Q (reaction force), and according to Newton's 3rd law, the two forces are equal and opposite.

The magnitude of the electric force between two single-point charges is

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the two charges

In this problem we have:

$q_1=+4.5\cdot 10^{-6}C$ is charge 1

$q_2=+7.2\cdot 10^{-6}C$ is charge 2

r = 0.30 cm = 0.003 m is the separation

So, the electric force between the two charges is

$F=(9\cdot 10^9)\frac{(4.5\cdot 10^{-6})(7.2\cdot 10^{-6})}{(0.003)^2}=32400 N$

The electric force between two charged objects has direction as follows:

- If the two objects have charges of opposite signs (+ and -), the force between them is attractive

- If the two objects have charges of same sign (++ or --), the force between them is repulsive

In this problem, the two charges are:

$q_1=+4.5\cdot 10^{-6}C$ is charge 1

$q_2=+7.2\cdot 10^{-6}C$ is charge 2

We see that the two charges have same sign: therefore, the force between them is repulsive.

The electric force between the proton and the electron in the atom can be written as

$F_E=k\frac{q_1 q_2}{r^2}$

where

$q_1 = q_2 = e = 1.6\cdot 10^{-19}C$ is the magnitude of the charge of the proton and of the electron

$r=5.3\cdot 10^{-11} m$ is the separation between them

So the force can be rewritten as

$F_E=\frac{ke^2}{r^2}$

The gravitational force between the proton and the electron can be written as

$F_G=G\frac{m_p m_e}{r^2}$

where

G is the gravitational constant

$m_p = 1.67\cdot 10^{-27}kg$ is the proton mass

$m_e=9.11\cdot 10^{-27}kg$ is the electron mass

Comparing the 2 forces,

$\frac{F_E}{F_G}=\frac{ke^2}{Gm_p m_e}=\frac{(9\cdot 10^9)(1.6\cdot 10^{-19})^2}{(6.67\cdot 10^{-11})(1.67\cdot 10^{-27})(9.11\cdot 10^{-31})}=2.3\cdot 10^{39}$

8 0

3 years ago